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can someone help me on a website for the classification of quadrics(ellipsoids,paraboloids,....) starting from the general second order equation.

Thanks

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- #1

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can someone help me on a website for the classification of quadrics(ellipsoids,paraboloids,....) starting from the general second order equation.

Thanks

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HallsofIvy

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mathwonk

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What I really mean is:

you have a general equation of the second order in three variables and by translations

and rotations to remove Xy-Xz and yz terms you transform it to the canonical forms

of quadrics.(ellipsoids,hyperboloids,cone......)

Bye

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mathwonk

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do you want to know how to complete the square in a quadratic equation? thats all it is.

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mathwonk

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in one variable you start from X^2 + bX + c, and you get

X^2 + bX + (b/2)^2 +c - (b/2)^2 = (X+ b/2)^2 + [(4c-b^2)]/4.

in 2 variables you start from X^2 + bXY + cY^2, and get

X^2 + bXY +(bY/2)^2 + cY^2 - (bY/2)^2

= (X+bY/2)^2 + [4c-b^2]/4 Y^2.

so now you replace X+ bY/2 by W and you have W^2 + [4c-b^2]/4 Y^2.

there are no WY terms, and so you can tell which quadric you have by the sign of

[4c-b^2]/4.

maybe you want to expand this a little, by throwing in a third letter Z, but this is the main trick.

i guess for a general homogeneous function of three variables you want to diagonalize the matrix representing the quadric.

the previous discussion above works for Z = quadratic in (X,Y), which arises in calculus of two variables.

X^2 + bX + (b/2)^2 +c - (b/2)^2 = (X+ b/2)^2 + [(4c-b^2)]/4.

in 2 variables you start from X^2 + bXY + cY^2, and get

X^2 + bXY +(bY/2)^2 + cY^2 - (bY/2)^2

= (X+bY/2)^2 + [4c-b^2]/4 Y^2.

so now you replace X+ bY/2 by W and you have W^2 + [4c-b^2]/4 Y^2.

there are no WY terms, and so you can tell which quadric you have by the sign of

[4c-b^2]/4.

maybe you want to expand this a little, by throwing in a third letter Z, but this is the main trick.

i guess for a general homogeneous function of three variables you want to diagonalize the matrix representing the quadric.

the previous discussion above works for Z = quadratic in (X,Y), which arises in calculus of two variables.

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mathwonk

yes that's.for three variables I think you have to diagonalize the matrix.First you you must

compute the eigen-values of the matrix .

Thank you for your help.

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